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Second shifting theorem of laplace transform

  1. Aug 26, 2014 #1
    1. why do we need to use shifted unit step function in defining second shifting theorem?
    2. why don't we instead calculate laplace transform of a time shifted function just by replacing t by t-a?
    3. everywhere in the books as well as internet i see second shifting theorem defined for f(t-a).u(t-a),
    why not just f(t-a)?
    4. what is the value of laplace transforms for negative limits?
     
  2. jcsd
  3. Aug 27, 2014 #2

    HallsofIvy

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    The Laplace transform is only defined for functions defined on [itex][0, \infty)[/itex] (since it requires an integral from 0 to [itex]\infty[/itex]). If f is defined on [itex][0, \infty)[/itex], f(t- a) is defined on [itex][a, \infty)[/itex]. Multiplying by u(t- a) just redefines the value to be 0 for [itex]0\le x< a[/itex] so that it is still defined on [itex][0, \infty)[/itex].

    I don't know what you mean by "negative limits". If you are referring to the limits of integration or the values on which the function is defined, as I said above, they must be 0 and [itex]\infty[/itex]. The "value of the Laplace transform" is simply not defined for any other values.
     
  4. Aug 27, 2014 #3
    My question is why do we use unit time shifted step function?
    I never said f is defined between zero and infinity.
    f can also have t- values.
    Then f(t-a) does not mean f is defined between a and infinity.
    f(t-a) just means, f(t) shifted by "a" time, where "a" can be negative, positive or zero.
    In case "a" is negative, its the advance time and the function gets shifted towards left (for t-).
     
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